Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1991 AJHSME Problems/Problem 7"

(→‎Solution)
(→‎See Also)
 
Line 31: Line 31:
  
 
==See Also==
 
==See Also==
−
 
−
{{AJHSME box|year=1991|num-b=6|num-a=8}}
 
−
[[Category:Introductory Algebra Problems]]
 
−
{{MAA Notice}}
 

Latest revision as of 15:34, 24 February 2024

Problem

The value of $\frac{(487,000)(12,027,300)+(9,621,001)(487,000)}{(19,367)(.05)}$ is closest to

$\text{(A)}\ 10,000,000 \qquad \text{(B)}\ 100,000,000 \qquad \text{(C)}\ 1,000,000,000 \qquad \text{(D)}\ 10,000,000,000 \qquad \text{(E)}\ 100,000,000,000$

Solution

We can make the approximations \begin{align*} 487,000 &\approx 500,000 \\ 12,027,300 &\approx 12,000,000 \\ 9,621,001 &\approx 10,000,000 \\ 19,367 &\approx 20,000. \end{align*}

Using these instead of the original numbers for an estimate, we have \begin{align*} \frac{(500,000)(12,000,000)+(10,000,000)(500,000)}{(20000)(.05)} &= \frac{500,000\times 22,000,000}{1000} \\ &= 500,000 \times 22,000 \\ &= 1.1\times 10^{10} \\ &\approx 10,000,000,000 \rightarrow \boxed{\text{D}}. \end{align*}


~*If you typed this into a calculator or actually got the answer, it would be AROUND 10 billion.*~

~*Calypso*~

"Google says the answer is 10,887,305,816.1" ~ xxsc

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AJHSME_Problems/Problem_7&oldid=216188"